Question

Consider the polynomial P(x)=x⁴-3x²-4.What are the solutions to x⁴-3x²-4=0.

Answer 1

The solutions are
`x = 2` and
`x = -2`.

Explanation:

We have the following equation:

`1)x^(4) - 3x^(2) - 4 = 0`.

The first step to solve this problem is using

`2) y = x^(2)`

We replace in the equation 1, find the values of y, and then we replace in equation 2) to find the values of x.

To solve the equations, it is important to know how we find the roots of a second order polynomial.

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = (x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

In this problem, we have

`x^(4) - 3x^(2) - 4 = 0`

`y = x^(2)`

So

`y^(2) - 3y - 4 = 0`

So:
`a = 1, b = -3, c = -4`

`\bigtriangleup = b^(2) - 4ac = (-3)^(2) -4(1)(-4) = 25`

`y_(1) = (-b + √(\bigtriangleup))/(2*a) = (3 + √(25))/(2) = 4`

`y_(2) = (-b - √(\bigtriangleup))/(2*a) = (3 - √(25))/(2) = -1`

The values of y are
`y_(1) = 4, y_(2) = -1`

We also have that:

`y = x^(2)`

So

`4 = x^(2)`

`x = \pm √(4)`

`x = \pm 2`

And

`-1 = x^(2)`

There is no real solution for this. So our only solutions are
`x = 2` and
`x = -2`.